Calculus of Real and Complex Variables

14.7 The Dual Space Of C0

(ℝp)

Consider the dual space of C0(ℝp) where ℝp is a polish space in which the balls have compact closure. It
will turn out to be a space of measures. To show this, the following lemma will be convenient. Recall
C0

(ℝp)

is defined as follows.

Definition 14.7.1f ∈ C0

(ℝp )

means that for every ε > 0 there exists a compact set K such that

|f (x)|

< ε whenever x

∕∈

K. Recall the norm on this space is

p
||f||∞ ≡ ||f|| ≡ sup{|f (x)| : x ∈ ℝ }

The next lemma has to do with extending functionals which are defined on nonnegative functions to
complex valued functions in such a way that the extended function is linear. This exact process was used
earlier with the abstract Lebesgue integral. Basically, you can do it when the functional “desires to be
linear”.

Let L ∈ C0(ℝp)′. Also denote by C0+(ℝp) the set of nonnegative continuous functions defined on
ℝp.

Definition 14.7.2Letting L ∈ C0

(ℝp)

′, define for f ∈ C0+(ℝp)

+ p
λ(f ) = sup{|Lg| : |g| ≤ f,g ∈ C 0 (ℝ )}.

Note that λ(f) < ∞ because

|Lg|

≤

∥L∥

∥g∥

≤

∥L∥

||f|| for

|g|

≤ f. Isn’t this a lot like the total
variation of a vector measure? Indeed it is, and the proof that λ wants to be linear is also
similar to the proof that the total variation is a measure. This is the content of the following
lemma.

Lemma 14.7.3If c ≥ 0,λ(cf) = cλ(f),f1≤ f2implies λ

(f1)

≤ λ

(f2)

, and

λ(f1 + f2) = λ(f1) + λ(f2).

Also

0 ≤ λ (f ) ≤ ∥L∥||f ||∞

Proof: The first two assertions are easy to see so consider the third. For i = 1,2 and fi∈ C0+

is the polar decomposition of the complex measure ν. Then with this convention, the above
representation is

∫ p
L (f) = p f dν, |ν|(ℝ ) = ||L ||.
ℝ

Also note that at most one ν can represent L. If there were two of them νi,i = 1,2, then ν1− ν2 would
represent 0 and so

|ν1 − ν2|

p
(ℝ )

= 0. Hence ν1 = ν2.

All of this works with virtually no change if ℝp is replaced with a Polish space, a complete separable
metric space in which closed balls are compact, or even more generally for a locally compact
Hausdorff space. In the first case, you simply replace ℝp with X, the name of the Polish space. In
the second case there are only a few technical details. However, I am trying to emphasize the
case which is of most interest and all of the principal ideas are present in this more specific
example.

The following is a rather important application of the above theory along with the material on Fourier
transforms presented earlier. It has to do with the fact that the characteristic function of a probability
measure is unique so if two such measures have the same one, then they are the same measure. This can
actually be shown rather easily. You don’t have to accept this kind of thing on faith and speculations based
on special cases.

Example 14.7.6Let μ and ν be two Radon probability measures on the Borel sets of ℝp. The typicalsituation is that these are probability distribution functions for two random variables.Then thecharacteristic function of μ is

(2π)

p∕2times the inverse Fourier transform of μ

∫
ϕμ (t) ≡ eit⋅xdμ (x)
ℝp

then a very important theorem from probability says that if ϕμ

(t)

= ϕν

(t)

, then the two measures areequal. This is very easy at this point, but not so easy if you don’t have the general treatment of Fouriertransforms presented above.

We have F−1

(μ)

= F−1

(ν)

in G∗. Therefore, μ = ν in G∗ and by definition, ∫ℝpψdμ = ∫ℝpψdν for all
ψ ∈G. But by the Stone Weierstrass theorem, G is dense in C0

(ℝp)

and so the equation holds for all
ψ ∈ C0

(ℝp )

. Now Cc

(ℝp )

⊆ C0

(ℝp )

and so the equation also holds for all ψ ∈ Cc

(ℝp)

. By
uniqueness in the Riesz representation theorem, it follows that μ = ν. You could also use Theorem
14.7.5.